232541184 Certain Numerical Problems Chemical Engineering MATLAB

5/21/2018 232541184 Certain Numerical Problems Chemical Engineering MATLAB

1/44

Exercise 1:

Write a program to convert a given temperature from F to C:

Solution:We will store a number in a variable. That number is the value of a temperature in

Fahrenheit. We will convert this value to centigrate. Enter the following commands.

>> tf=45;

>> tc=(tf-32)*5/9

the result will be

tc =

7.2222

Exercise 2:

Write a program to calculate the vapor pressure of water according to Antoine equation:

Po=exp(A-B/(T+C))

Where T is any given temperature in Kelvin and A, B, and C are Antoine coefficients:

A=18.3036

B=3816.44

C= -46.13

Solution:

Let temperature equal to 373.15 k, write the following code.

>> T=373.15;

>> A=18.3036;

>> B=3816.44;

>> C= -46.13;

>> Pw=exp(A-B/(T+C))

Pw =759.9430

Note: you can use any variable name in your code.


2/44

15% X

25% S

40% T

20% Z

D=80

B=?

F=100kg

?% X

?% S

?% T

?% Z

15% X

25% S

40% T20% Z

Exercise 3:

For the following distillation column write a code to find the value of stream B and the

compositions of stream D?Solution:

Type the commands as m-file. Then copy it to command window.

F=100;

D=80;

B=F-D

XF=0.15;

SF=0.25;

TF=0.4;ZF=0.2;

XB=0.15;

SB=0.25;

TB=0.4;

ZB=0.2;

XD=(F*XF-B*XB)/D*100

SD=(F*SF-B*SB)/D*100

TD=(F*TF-B*TB)/D*100

ZD=(F*ZF-B*ZB)/D*100

Then after pressing enter the result will be:

B =

20

XD =

15

SD =25

TD =

40

ZD =

20


3/44

Exercise 1:

Find the slope of a tangent line tox+3x-2at x=2.

Solution:

To find the slope of a tangent line tox+3x-2 at point 2, we need to find the derivative

and to evaluate it atx=2.

First we find the derivative of the function;

>> diff(x^2+3*x-2)

ans =

2*x+3

Then we representative the derivative as inline function

>>f = inl ine('2*x+3')

f = Inline funct ion: f(x) = 2*x+3

And, finally, we evaluate the derivative at 2

>> f(2) ans = 7


4/44

Solution:

By making component material balance on each component within the mixer you can

reach to a system of three equations which can be solve by using the command solve to find

the unknowns A, B, C.

Type the following command:

[A,B,C]=solve('.5*A+.3*B=.4*C','.2*A+.3*B=.2*C','.3*A+.4*B+100=.4*C')The results will be:

A =

600.

B =

200.

C =

900.

A= ?

50% Xylene

20% Toluene

30% Benzene

Exercise 2:

For the mixer shown below write a code to find the values of streams A, B and C?

W= 100 Kg/hr100% Benzene

B= ?

30% Xylene

30% Toluene

40% Benzene

C= ?

40% Xylene20% Toluene

40% Benzene


5/44

27

Type the same command by entering w=100 as the fourth equation:

>>[A,B,C,W]=solve('.5*A+.3*B=.4*C','.2*A+.3*B=.2*C','.3*A+.4*B+W=.4*C','W

=100')The results will be:

A =

600.

B =

200.

C =

900.

W =100.

Practice problems:

1. Factorx+3xy+3xy+y.

2. Simplify (x-8)/(x-2).

3. Expand (x+1)(x-5)(2x+3).

4. Solve sinx = 2-x forx.

5. Solve 5x+2y+4z = 8, -3x+y+2z = -7, 2x+y+z = 3 forx, y andz.

6. Solvey-5xy-y+6x+x = 2 forx.

7. Find the first derivative of the function (sinx/(ln(x2+1))-e

xand evaluate it at x=3.

8. Find the 12th derivative of the function (x/2+1)65

9. Find the first and second partial derivatives of the function ex2

sinxy


6/44

Exercise 1:

Estimate the average boiling point of a Water/Ethanol system at different water

compositions (0, 0.1, 0.2 ,0.3.1.0) knowing that.Boiling point of water =100

oC

Boiling point of Ethanol =78.35oC

Mixture boiling point= Xwater TBpwater+ Xethanol TBpethanol

Solution:

Write the code as m-file then copy it to command window.

TBPwater=100;

TBPethanol=78.35;

Xwater=0:.1:1

Xethanol=1-Xwater

T=TBPwater*Xwater+TBPethanol*Xethanol

Then press enter, then the output will be:

Xwater =

0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000

0.8000 0.9000 1.0000

Xethanol =1.0000 0.9000 0.8000 0.7000 0.6000 0.5000 0.4000 0.3000

0.2000 0.1000 0

T =

78.3500 80.5150 82.6800 84.8450 87.0100 89.1750 91.3400 93.5050

95.6700 97.8350 100.0000

Exercise 2:

Calculate the Reynold Number at two different diameters D=0.2 and 0.5 m, using

different velocity's as u=0.1,0.2,0.3 1 m/s knowing that:

Re= (ud)/ , =0.001 , =1000

Solution:

Type the code

d1=0.2;

d2=0.5;


7/44

u=0.1:.1:1;

m=0.001;

p=1000;Re1=u*p*d1/m

Re2=u*p*d2/m

The results will be

Re1 =

1.0e+005 *

0.2000 0.4000 0.6000 0.8000 1.0000 1.2000 1.4000 1.6000

1.8000 2.0000

Re2 =

1.0e+005 *

0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000

4.5000 5.0000

Practice Problems

1. Create a vector of the even whole numbers between 31 and 75.

2. Let x = [2 5 1 6].

a. Add 16 to each element.

b.

Add 3 to just the odd-index elements.

c. Compute the square root of each element.

d. Compute the square of each element.

3. Let x = [3 2 6 8]' and y = [4 1 3 5]' .

a. Add the elements in x to y

b. Raise each element of x to the power specified by the corresponding element in y.

c. Divide each element of y by the corresponding element in x

d. Multiply each element in x by the corresponding element in y, calling the result "z".


8/44

Exercise 1:

Start with a fresh M-file editing window. Write a code to convert the temperature in

Celsius into F and then into R for every temperature from 0 increasing 15 to 100C.

Combine the three results into one matrix and display them as a table.

SolutionWrite the code

tc = [0:15:100]; % tc is temperature Celsius, tf is temp deg F,

tf = 1.8.*tc + 32; % and tr is temp deg Rankin.

tr = tf + 459.69;

t = [tc',tf',tr'] % combine answer into one matrix

The results will be

t =

0 32.0000 491.6900

15.0000 59.0000 518.6900

30.0000 86.0000 545.6900

45.0000 113.0000 572.6900

60.0000 140.0000 599.6900

75.0000 167.0000 626.6900

90.0000 194.0000 653.6900


9/44

Exercise 2:

Estimate the average boiling point of a Water/Ethanol system at different water

compositions (0, 0.1, 0.2 ,0.3.1.0) knowing that.Boiling point of water =100

oC

Boiling point of Ethanol =78.35oC

Mixture boiling point= Xwater TBpwater+ Xethanol TBpethanol

Note: Make a matrix of three columns; first and second for water and ethanol

compositions and the third for average density.

Solution:

Write the codeWD=1000;

ED=789;

Xw=0:.1:1;

Xe=1:-.1:0;

AD=WD*Xw+ED*Xe;

A=Xw';

A(:,2)=Xe';

A(:,3)=AD';A

The output of the above code will be

A =

1.0e+003 *

0 0.0010 0.7890

0.0001 0.0009 0.8101

0.0002 0.0008 0.8312

0.0003 0.0007 0.8523

0.0004 0.0006 0.8734

0.0005 0.0005 0.8945

0.0006 0.0004 0.9156

0.0007 0.0003 0.9367

0.0008 0.0002 0.9578

0.0009 0.0001 0.9789

0.0010 0 1.0000


10/44

Exercise 3:

Make a table of (x vs. y) values for the three component ethanol, water and benzene a

T=373.15 k. Knowing that the vapor pressure of these three components are calculated by:Ethanol P

oe=exp(18.5242-3578.91/(T-50.5))

Water Pow=exp(18.3036-3816.44/(T-46.13))

Benzene Pob=exp(15.9008-2788.51/(T-52.36))

Where

Ki= Poi /Pt, Pt=760, yi =Kixi

Solution:

T=373.15;

Pe=exp(18.5242-3578.91/(T-50.5));

Pw=exp(18.3036-3816.44/(T-46.13));

Pb=exp(15.9008-2788.51/(T-52.36));

Ke=Pe/760; Kw=Pw/760; Kb=Pb/760;

Xe=0:.1:1; Ye=Ke*Xe;

Xw=0:.1:1; Yw=Kw*Xw;

Xb=0:.1:1; Yb=Kb*Xb;

A=[Xe',Ye' ,Xw',Yw',Xb',Yb']The output of the above code will be:

A=

0 0 0 0 0 0

0.1000 0.2223 0.1000 0.1000 0.1000 0.1777

0.2000 0.4445 0.2000 0.2000 0.2000 0.3554

0.3000 0.6668 0.3000 0.3000 0.3000 0.5331

0.4000 0.8890 0.4000 0.4000 0.4000 0.7107

0.5000 1.1113 0.5000 0.5000 0.5000 0.8884

0.6000 1.3335 0.6000 0.6000 0.6000 1.0661

0.7000 1.5558 0.7000 0.6999 0.7000 1.2438

0.8000 1.7780 0.8000 0.7999 0.8000 1.4215

0.9000 2.0003 0.9000 0.8999 0.9000 1.5992

1.0000 2.2225 1.0000 0.9999 1.0000 1.7769


11/44

Example:solve the following system equations:x + 2y = 5

2x + 6y = 14

To solve these equations write the code;

a=[1 2; 2 6];

b=[5; 14];

z=a\b

yields

z =

1.0000

2.0000

this results means that x=z(1)=1 and y=z(2)=2

Example:Solve the system of linear equations given below using Matlab.

3.5u+2v =5

-1.5u+2.8v+1.9w=-1-2.5v+3w=2

Solution

A=[3.5 2 0;-1.5 2.8 1.9;0 -2.5 3];

B=[5;-1;2]

X=A\B

X=

1.4421

-0.0236

0.6470


12/44

Exercise 1:

For the following separation system, we know the inlet mass flowrate (in Kg/hr) and

the mass fractions of each species in the inlet (stream 1) and each outlet (streams 2, 4, and

5). We want to calculate the unknown mass flow rates of each outlet stream.

Solution:

If we define the unknowns as

x1=F1, x2=F2 , x3=F3

and set up the mass balances for

1. the total mass flow rate

x1 +x2 +x3 =10

2. the mass balance on species 10.04x1+0.54x2+0.26x3=0.2*10

3. the mass balance on species 2

0.93x1+0.24X2 =0.6*10

These three equations can be written in matrix form

=

6

2

10

3

2

1

024.093.0

26.054.004.0

111

x

x

x

To find the values of unknown flowrates write the code:

A=[1, 1, 1; .04, .54, .26; .93, .24 ,0];

B=[10; .2*10; .6*10];

X=A\B;

F1=X(1),F2=X(2),F3=X(3)

The results will be

F1 =5.8238

F2 =

2.4330

F3 =

1.7433

F=10

A=20%B=60%

C=20%

F3=?

A=26%

B=0 %

C=74%

F2=?

A=54%B=24%

C=22%

F1=?

A=4 %

B=93%

C=3 %


13/44

Exercise 2:

Fit the following data describing the accumulation of species A over time to a second

order polynomial, then by using this polynomial, predict the accumulation at 15 hours.Time (hr) 1 3 5 7 8 10

Mass A acc. 9 55 141 267 345 531

Solution:

First, input the data into vectors, let:

a = [9, 55, 141, 267, 345, 531];

time = [1, 3, 5, 7, 8, 10];

Now fit the data using polyfit

coeff = polyfit(time,a,2)

coeff =

5.0000 3.0000 1.0000

So, Mass A = 5*(time)2+ 3 * (time) + 1

Therefore to calculate the mass A at 15 hours

MApred = polyval(coeff,15)

MApred =

1.1710e+003

Exercise 3:

Use matlab to fit the following vapor pressure vs temperature data in fourth order

polynomial. Then calculate the vapor pressure when T=100 OC.

Temp (C) -36.7 -19.6 -11.5 -2.6 7.6 15.4 26.1 42.2 60.6 80.1

Pre. (kPa) 1 5 10 20 40 60 100 200 400 760

Solution:

vp = [ 1, 5, 10, 20, 40, 60, 100, 200, 400, 760];

T = [-36.7, -19.6, -11.5, -2.6, 7.6, 15.4, 26.1, 42.2, 60.6, 80.1];


14/44

p=polyfit(T,vp,4)

%evaluate the polynomial at T=100

pre= polyval(p,100)

the results will be

p =

0.0000 0.0004 0.0360 1.6062 24.6788

pre =

1.3552e+003

PRACTICE PROBLEMS1) Solve each of these systems of equations by using the matrix inverse ( Gaussian

Elimination ) method.

A) -2x1+x2=-3

-2x1+x2=3.00001

B) x1+ x2+ x3= 4

2x1+ x2+ 3x3= 7

3x1+ x2+6x3 = 2

C ) x1+2x2-x3=3

3x1-x2+2x3=1

2x1-2x2+3x3=2

x1-x2+x4=-1

D) 5x1+3x2+7x3-4=03x1+26x2-2x3-9=0

7x1+2x2+10x3-5=0

2. Find the solution to

r + s + t + w = 4

2r - s + w = 2

3r + s - t - w = 2

r - 2s - 3t + w = -3

using the matrix method.


15/44

3. Find the solution to

2x1+ x2- 4x3+ 6x4+ 3x5- 2x6= 16

-x1+ 2x2+ 3x3+ 5x4- 2x5 = -7

x1- 2x2- 5x3+ 3x4+ 2x5+ x6= 1

4x1+ 3x2- 2x3+ 2x4 + x6= -1

3x1 + x2- x3+ 4x4+ 3x5+ 6x6= -11

5x1+ 2x2- 2x3+ 3x4+ x5+ x6= 5

using the matrix method.

4. Find the 3rd

order polynomial that satisfies the above data of water for saturation

temperature and pressure. Using the predicted polynomial, compute the saturation pressureat 65 C.

Temp

(C)

0 10 20 30 40 50 60 70 80 90 100

Pre.

(kPa)

.6108 1.227 2.337 4.241 7.375 12.335 19.92 31.16 47.36 70.11 101.33


16/44

Exercise 1:

Write a program to calculate average density, conductivity and specific heat for water in

the range of temperatures from 0 to 50 C. Knowing that this parameters for water are afunction of temperature such as the following equations.

The Density

= 1200.92 1.0056 TKo+ 0.001084 * (TK

o)

2

The condactivity

K= 0.34 + 9.278 * 10-4

.TKo

The Specific heat

CP= 0.015539 (TKo 308.2)

2+ 4180.9

Note: take 11 point of temperatures

Solution:

ap=0;

aKc=0;

aCp=0;

for T=273:5:323

p= 1200.92 - 1.0056*T+ 0.001084 * T^2;

Kc = 0.34 + 9.278 * 10^-4 *T;

Cp = 0.015539*(T - 308.2)^2 + 4180.9;

ap=ap+p;

aKc=aKc+Kc;

aCp=aCp+Cp;

end

Averagedensity=ap/11

Averageconductiv ity=aKc/11

Averagespecif icheat=aCp/11Gives the results

Averagedensi ty =

997.7857

Averageconduct ivi ty =

0.6165

Averagespecificheat =

4.1864e+003


17/44

Exercise 2:

Use loop to find the bubble point of ternary system (Ethanol 40 mol%, Water 20 mol%

and Benzene 40 mol%). Knowing that the vapor pressure for three components are calculatedby:

Ethanol Poe=exp(18.5242-3578.91/(T-50.5))

Water Pow=exp(18.3036-3816.44/(T-46.13))

Benzene Pob=exp(15.9008-2788.51/(T-52.36))

Where

Ki= Poi /Pt

Pt=760

yi =Kixi

At Bubble point yi=Kixi =1

Solution:

Xe=.4;Xw=.2;Xb=.4;

for T=273.15:.01:450;

Pe=exp(18.5242-3578.91/(T-50.5)); % Vapor pressure of Ethanol

Pw=exp(18.3036-3816.44/(T-46.13)); % Vapor pressure of Water

Pb=exp(15.9008-2788.51/(T-52.36)); % Vapor pressure of Benzene

% evaluation of equilibrium constants

Ke=Pe/760; % Ethanol

Kw=Pw/760; % Water

Kb=Pb/760; % Benzene

% The condition to find the boiling point

sum=Ke*Xe+Kw*Xw+Kb*Xb;

if sum>1

breakend

end

T

Gives the result

T =

355.5300


18/44

Exercise 3:

Given that the vapor pressure of methyl chloride at 333.15 K is 13.76 bar, write a code

to calculate the molar volume of saturated vapor at these conditions using Redlich/Kwongequation. Knowing;

a=0.42748*R2Tc

2.5/Pc

b=0.08664*RTC/PC

Vi+1=(RT/P)+b- (a*(Vi-b))/(T1/2

PVi(Vi+b))

R=83.14 , Tc= 416.3 k , Pc= 66.8 bar

Solution:

T=333.15; Tc=416.3;

P=13.76; Pc=66.8; R=83.14;a=0.42748*R^2*Tc^2.5/Pc;

b=0.08664*R*Tc/Pc;

V(1)=R*T/P;

for i=1:1:10

V(i+1)=(R*T/P)+b-(a*(V(i)-b))/(T^.5*P*V(i)*(V(i)+b));

end

V'

The program results as follows

ans =

1.0e+003 *

2.0129

1.7619

1.7219

1.7145

1.71311.7129

1.7128

1.7128

1.7128

1.7128

1.7128

Note that after the 6 th trail the value of Vi is constant and further trails does not have

any effect on its value.


19/44

Practice problems:

1/ evaluate the output of a given MATLAB code for each of the cases indicated below.

if z < 5w = 2*z

elseif z < 10

w = 9 - z

elsei f z < 100

w = sqrt(z)

else

w = z

end

a. z = 1 w = ?

b. z = 9 w = ?

c. z = 60 w = ?

d. z = 200 w = ?

2/ Write brief code to evaluate the following functions.

t = 200 when y is below 10,000

= 200 + 0.1 (y - 10,000) when y is between 10,000 and 20,000

= 1,200 + 0.15 (y - 20,000) when y is between 20,000 and 50,000

= 5,700 + 0.25 (y - 50,000) when y is above 50,000

Test cases:

a. y = 5,000 t = ?

b. y = 17,000 t = ?b. y = 25,000 t = ?

c. y = 75,000 t = ?


20/44

Exercise 1:

The laminar velocity profile in a pipe, can be given in equation Vx=Vmax(1-(2*r/di)2)

Where

r= distance from pipe center

di: Inner diameter of pipe (di=0.2 m)

Vmax: maximum velocity in pipe (Vmax=2 m)

Plot the velocity profile in a pipe?

SolutionWrite the following code

di=.2; Vmax=2

r=-.1:.001:.1

Vx=Vmax*(1-((2*r)/di).^2)

plot(Vx,r)

xlabel('Velocity (m/s)')

ylabel('Distance from pipe center (m)')The result will be the figure.


21/44

Exercise 2:

Plot k as a function of temperature from 373 to 600 k (use 50 point of Temperature).

If you know:

K=A exp(B-(C/T)-ln(D*T)-E*T+F*T2)

where

A=3.33

B=13.148

C=5639.5

D=1.077

E=5.44*10-4

F=1.125*10-7

Solution

Write the following code

A=3.33; B=13.148; C=5639.5; D=1.077; E=5.44e-4; F=1.125e-7;Ts=(600-373)/49;%Ts means temperature increment

m=0;

for T=373:Ts:600

m=m+1

Temp(m)=T

K(m)=A*exp(B-(C/T)-log(D*T)-E*T+F*T^2)

end

plot(Temp, K)

xlabel('Temperature')

ylabel('K value')

The result will be the figure.


22/44

Liquid x,T

Vapor y,T

Exercise 3:

Plot Txy diagram for ethanol/water system. Knowing that the vapor pressure for three

components are calculated by:

Ethanol Poe=exp(18.5242-3578.91/(T-50.5))

Water Pow=exp(18.3036-3816.44/(T-46.13))

Where

Ki= Poi /Pt

Pt=760

yi =Kixi

At Bubble point yi=Kixi =1

Solution:

Xe=0:.1:1;

Xw=1-Xe;

for m=1:11

for T=273.15:.01:450;

Pe=exp(18.5242-3578.91/(T-50.5)); % Vapor pressure of Ethanol

Pw=exp(18.3036-3816.44/(T-46.13)); % Vapor pressure of Water

Ke=Pe/760; % Ethanol


23/44

Kw=Pw/760; % Water

sum=Ke*Xe(m)+Kw*Xw(m);

if sum>1break

end

end

Temp(m)=T;

ye(m)=Ke*Xe(m);

end

plot(Xe,Temp,'k-+',ye,Temp,'k-*')

axis ([0 1 340 380])

xlabel('x,y for ethanol ')

ylabel('Temperatre K')

title('Txy diagram of ethanol/ water system')


24/44

Practice Problems

1. Plot the log of the values from 1 to 100.2. Plot the parametric curve x = t cos(t), y = t sin(t) for 0 < t < 10.

3. Plot the cubic curve y = x. Label the axis and put a title "a cubic curve" on the top of

the graph.

4. Plot y=(x3+x+1)/x, for 4< x


25/44

Practice Problems

Q1:write program to calculate the volumetric and mass flow rate of a liquid flowing in a pipewith a velocity equal to 0.5 m/s. Knowing that the diameter of this pipe is 0.1 m and thedensity of this liquid is 890 kg/m3 ?

Solution:

d=0.1;p=890;u=.5;

A=(pi/4)*d^2;

Volflow=u*A

Massflow=Volflow*p

A =

0.0079

Volflow =

0.0039

Massflow =

3.4950

Q2:The heat capacity of ethylene is given by the equation:3925 T1017.6T108.30.156T3.95Cp ++=

Write a program to evaluate the heat capacity T=40 C Then claculate the ethylene enthalpy(h) when T1=80 in which.H=Cp(T1-Tref)Tref=0 CSolution:

T=40;

Tref=0;

T1=80;Cp=3.95+.159*T-8.3*10^-7*T^2+17.6*10^-9*T^3

h=Cp*(T1-Tref)

Cp =

10.3098

h =

824.7839


26/44

ATCost = 2000

Q3:Write a program to calculate the cost of iron used in manufacturing a cylindrical tank,where.

DLDA +=2

5.0

in which:T: Thickness= 0.01 mD: Diameter=2 mA: Surface areaL: length = 5 mCost in $ dollarSolution:

T=0.01;

L=5;

D=2;

A=0.5*pi*D^2+pi*D*L;

% Cost in $ dollar

Cost=2000*T*A

Cost =

753.9822

Q4:Calculate the heat required to increase the temperature of 1 mol of methane from 533.15K to 873.15 C in a flow process at a pressure approximately 1 bar. where

22 +++= DTCTBTAR

Cp

A=1.702 , B=9.081*10-3

, C=-2.164*10-6

, D=0R=8.314and

=Tout

Tin

CpdtnQ

Solution:

syms T;

T1=533.15;

T2=873.15;

A=1.702;

B=9.081e-3;

C=-2.164e-6;


27/44

R=8.314;

Cp=(A+B*T+C*T^2);

Q=R*int(Cp,T1,T2)Q =

1.9778e+004

Q5:For the following distillation column calculate the values of F1, F3 and F4?

solution:

[F1,F3,F4]=solve('.2*F1+250=.5*F3+.2*F4','.3*F1+250=.3*F3+.4*F4','.

5*F1=.2*F3+.4*F4')

F1 =

1000

F3 =

500

F4=

1000

F1=?

20%A

30%B

50%D

F3=?

50%A

30%B20%D

F4=?

20%A

40%B

40%D

F2=500

50%A

50%B


28/44

Q6:Estimate the average density of a Water/Ethanol mixture at different water compositionsknowing that.Water density =1000 kg/m3Ethanol density =780 kg/m3Mixture density= Xwater Water density + Xethanol Ethanol densitySolution:

Pwater=1000;

Pethanol=780;

Xwater=0:.1:1

Xethanol=1-Xwater

Pav=Pwater*Xwater+Pethanol*Xethanol

Xwater =0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000

0.7000 0.8000 0.9000 1.0000

Xethanol =

1.0000 0.9000 0.8000 0.7000 0.6000 0.5000 0.4000

0.3000 0.2000 0.1000 0

Pav =

780 802 824 846 868 890 912

934 956 978 1000

Q7:Write a program to calculate the vapor pressure for the following system and to calculateand tabulate its k-values.P1=exp(14.2724-2945.47/(T+224))P2= exp(14.2043-2972.64/(T+209))Temperature range =75-125 C

oUse 11 point for temperature

Pt=101.3Ki= Pi /Pt

Solution:T=75:5:125;

P1=exp(14.2724-2945.47./(T+224))

P2= exp(14.2043-2972.64./(T+209))

Pt=101.3

K1= P1/PtK2= P2/PtA=[T,P1,P2,K1,K2]


29/44

Q8: Write a program to estimate the physical properties for water in the range oftemperatures from 273 to 323 K.The Density

= 1200.92 1.0056 T + 0.001084 T2The conductivityK= 0.34 + 9.278 * 10

-4T

The Specific heatCP= 0.015539 (T 308.2)

2+ 4180.9

Note: take 101 points of temperaturesSolution:

T=273:.5:323;

%The Density

P = 1200.92 - 1.0056*T + 0.001084*T.^2 ;

%The conductivity

K= 0.34 + 9.278* 10^-4*T;

%The Specific heat

CP = 0.015539*(T-308.2).^2 + 4180.9;

A=[T',P',K',CP']

Q9:Solve the following system of equations forx, y and zby using matrix method.

10x+3y+5z = 88x+2y+2z = 73x+4y-z = 6 .

Solution:

A=[10,3,5;8,2,2;3,4,-1];

B=[8;7;6];

X=A\B;

x=X(1),y=X(2),z=X(3)

Gives the resultsx =

0.7917

y =

0.7917

z =

-0.4583


30/44

Q10: Find the solution to this system of equationsr + s + t + w = 4

2r - s + w = 23r + s - t - w = 2r - 2s - 3t + w = -3

using the matrix method.Solution:

A=[ 1 1 1 1;2 -1 0 1;3 1 -1 -1; 1 -2 -3 1];

B=[4;2;2;-3];

X=A\B;

r=X(1),s=X(2),t=X(3),w=X(4)

r =1

s =

1

t =

1

w =

1

Q11:Make a table of (x vs. y) values for the three component ethanol, water and benzene aT=373.15 k. Knowing that the vapor pressure of these three components are calculated by:

Note: take 11 point for each component form 0 to 1Ethanol P

oe=exp(18.5242-3578.91/(T-50.5))

Water Po

w=exp(18.3036-3816.44/(T-46.13))Benzene Pob=exp(15.9008-2788.51/(T-52.36))WhereKi= P

oi /Pt, Pt=760, yi =Kixi

Solution:T=373.15

Pe=exp(18.5242-3578.91/(T-50.5));

Pw=exp(18.3036-3816.44/(T-46.13));

Pb=exp(15.9008-2788.51/(T-52.36));

Ke=Pe/760;

Kw=Pw/760;

Kb=Pb/760;

Xe=0:.1:1;


31/44

Ye=Ke*Xe;

Xw=0:.1:1;

Yw=Kw*Xw;Xb=0:.1:1;

Yb=Kb*Xb;

A=[Xe',Ye',Xw',Yw',Xb',Yb']

A =

Xe Ye Xw Yw Xb Yb

0 0 0 0 0 0

0.1000 0.2223 0.1000 0.1000 0.1000 0.1777

0.2000 0.4445 0.2000 0.2000 0.2000 0.3554

0.3000 0.6668 0.3000 0.3000 0.3000 0.5331

0.4000 0.8890 0.4000 0.4000 0.4000 0.7107

0.5000 1.1113 0.5000 0.5000 0.5000 0.8884

0.6000 1.3335 0.6000 0.6000 0.6000 1.0661

0.7000 1.5558 0.7000 0.6999 0.7000 1.2438

0.8000 1.7780 0.8000 0.7999 0.8000 1.4215

0.9000 2.0003 0.9000 0.8999 0.9000 1.5992

1.0000 2.2225 1.0000 0.9999 1.0000 1.7769

Q12:Calculate the pressure of and ideal gas (with two digits accuracy) if the temperature ofideal gas is calculated from the equation;T=A*T^.3+B*T^.7A= 0.452B= 0.736Where this gas is follow ideal gas lawPV=RTV=1R=8.314Solution:

clear all

clc

A= 0.452; B= 0.736; V=1; R=8.314;

T(1)=273.15;


32/44

P(1)=R*T(1)/V

for m=1:100

T(m+1)=A*T(m)^.3+B*T(m)^.7;P(m+1)=R*T(m+1)/V ;

if abs(P(m+1)-P(m))


33/44

Note: calculate the volumetric flow rate to four digit accuracySolution:

A=0.35; B=0.98; C=0.342;

di=0.2;

u=1;%any initial assumption

Area=(pi/4)*di^2

Q(1)=u*Area

for i=1:100

u=A*u^.3+B*u^.6+C*u^.9;

Q(i+1)= Area* u

if abs(Q(i+1)-Q(i))


34/44

Q14:For the vapor liuqid seperater shown in fig. Write a program to calculate the values ofXA,XB,YA,YB, L and VIf you know:XA+XB=1YA+YB=1YA=KA*XA=2XAYB=KB*XB=2XB

Solution:

A=[1,1,0,0;0,0,1,1;-2,0,1,0;0,-.5,0,1];B=[1;1;0;0];

X=A\B;

xa=X(1),xb=X(2),ya=X(3),yb=X(4)

a=[xa,ya;xb,yb];

b=[.4*100;.6*100];

x=a\b;

L=x(1),V=x(2)

Gives the results

xa =

0.3333

xb =

0.6667

ya =

0.6667

yb =

0.3333

L =

80

V =

20.0000

Q15:Xylene, styrene, toluene and benzene are to be separated with the array of distillationcolumns that is shown below. Write a program to calculate the amount of the streams D, B,

D1, B1, D2 and B2 also to calculate the composition of streams D and B.

liquid

F=100 Kg/hr

40% A

60% B

Vapor

V=?

YA=?YB=?

L=?

XA=?

XB=?


35/44

SolutionBy making material balances on individual components on the overall separation train yieldthe equation setXylene: 0.07D1+ 0.18B1+ 0.15D2+ 0.24B2= 0.15 70Styrene: 0.04D1+ 0.24B1+ 0.10D2+ 0.65B2= 0.25 70Toluene: 0.54D1+ 0.42B1+ 0.54D2+ 0.10B2= 0.40 70

Benzene: 0.35D1+ 0.16B1+ 0.21D2+ 0.01B2= 0.20 70Overall material balances and individual component balances on column 2 can be used todetermine the molar flow rate and mole fractions from the equation of stream D.Molar Flow Rates: D = D1 + B1Xylene: XDxD = 0.07D1 + 0.18B1Styrene: XDsD = 0.04D1 + 0.24B1Toluene: XDtD = 0.54D1 + 0.42B1Benzene: XDbD = 0.35D1 + 0.16B1where

XDx = mole fraction of Xylene.


36/44

XDs = mole fraction of Styrene.XDt = mole fraction of Toluene.XDb =mole fraction of Benzene.Similarly, overall balances and individual component balances on column 3 can be used todetermine the molar flow rate and mole fractions of stream B from the equation set.Molar Flow Rates: B = D2 + B2Xylene: XBxB = 0.15D2 + 0.24B2Styrene: XBsB = 0.10D2 + 0.65B2Toluene: XBtB = 0.54D2 + 0.10B2Benzene: XBbB = 0.21D2 + 0.01B2where F, D, B, D1, B1, D2 and B2 are the molar flow rates in mol/min.

Solution;clear all

clc

A=[0.07, 0.18, 0.15, 0.24; 0.04, 0.24, 0.10, 0.65; 0.54, 0.42, 0.54, 0.1;0

.35, 0.16, 0.21, 0.01];

B=[0.15*70; 0.25*70; 0.4*70; 0.2*70];

X=A\B;

D1=X(1),B1=X(2),D2=X(3),B2=X(4),

D=D1+B1B=D2+B2

XDx=(.07*D1+.18*B1)/D

XDs=(.04*D1+.24*B1)/D

XDt=(.54*D1+.42*B1)/D

XDb=(.35*D1+.16*B1)/D

XBx=(.15*D2+.24*B2)/B

XBs=(.1*D2+.65*B2)/B

XBt=(.54*D2+.1*B2)/B

XBb=(.21*D2+.01*B2)/B

The results will be

D1 =

26.2500

B1 =

17.5000

D2 =


37/44

8.7500

B2 =

17.5000D =

43.7500

B =

26.2500

XDx =

0.1140

XDs =

0.1200

XDt =

0.4920

XDb =

0.2740

XBx =

0.2100

XBs =

0.4667

XBt =

0.2467

XBb =

0.0767

Q16:The experiemental values calculated for the heat capacity of ammonia from 0 to 500 are

T (C) Cp ( cal /g.mol C)

0 8.37118 8.472

25 8.514

100 9.035

200 9.824

300 10.606

400 11.347

500 12.045

1. Fit the data for the following function


38/44

32DTCTbTaCp +++=

Where T is in C

2. Then calculate amount of heat Q required to increase the temperature of 150 mol/hr ofammonia vapor from 0 C to 200 C if you know that:

=Tout

Tin

CpdtnQ

Solution:

T=[0,18,25,100,200,300,400,500]

Cp=[8.371, 8.472, 8.514, 9.035, 9.824, 10.606, 11.347, 12.045]

P=polyfit(T,Cp,3)n=150;

syms t

Cpf=P(4)+P(3)*t+P(2)*t^2+P(1)*t^3;

Q= n*int(Cpf, 0,200)

2.7180e+005

Q17: A simple force balance on a spherical particle reaching terminal velocity in a fluid isgiven by;Vt=((4g(p- )Dp)/(3CDp))^.5WhereVt : Terminalvelocity in m/sg : Acceleration of gravity

pp: Particle densityDp: The diameter of the spherical particle in mCD: Dimensionless drag coefficient.The drag coefficient on a spherical particle at terminal velocity varies with Reynolds number(Re) as followings:

CD=24/Re for Re< 0.1CD=24*(1+0.14* Re^0.7)/Re for 0.1=


39/44

Calculate the terminal velocity of spherical particle?Solution:The above problem can not be solved without using trail and error method. Therefore youmust assume a velocity to calculate the CD which is a important to calculate new velocity.Write the following code:

g=9.80665; p=994.6; pp=1800; mu=8.931e-4; Dp=0.000208;

% Assume vt of any initial terminal velocity for first trail

vt=1;

Velocity(1)=vt;

for m=1:1:20

Re=(Dp*vt*p)/mu;

if Re < 0.1

CD=24/Re;

elseif Re >= 0.1 & Re 1000 & Re 350000

CD=0.19-8*10^4/Re;

end

vt=((4*g*(pp-p)*Dp)/(3*CD*p))^.5;

Velocity(m+1)=vt;

if abs (Velocity(m+1)-Velocity(m))


40/44

0.0160

0.0159

0.01580.0158

0.0158

Q18: Unsteady-state ball loss 10% of its last height after each strike in ground. If this ball hasinitial height equal to 2 m and stop when its height reaches lower than 0.2 m. Plot its heightas a function of strike no.Solution:

clear all ,clc

h(1)=2;

m=1

while h(m)> .2

h(m+1)=.9*h(m);

m=m+1

end

plot(1:m,h,'k')

xlabel ('Strike No.')

ylabel('Ball Height m')


41/44

Q19:a tree raise in different raising rate, when its initial height is 5 meter write a program toplot a tree height as a function of time within ten years.Raising rate = (1/time).Solution:

clear all, clc

m=1

L(1)=5

Time(1)=0

for tim=1:10

L(m+1)=L(m)+(1/tim)

Time(m+1)=tim

m=m+1

end

plot(Time,L,k)

xlabel ('Time (year)')

ylabel('Tree height m')


42/44

Q20: A naphthalene sphere loss each day amount of weight equal to 0.2*(survace area)^0.5.If its initial diameter is 2 cm. write a program to plot naphthalene diameter, as a function oftime until the diameter is reach 0.2 m.

Note:Naphthalene density=0.890 gm/cm3

A=d2

V=(4/3) d3

Solution:

clear all

clc

p=.89;

m=1;ti=0;

time(1)=0

d(1)=2;

V(1)=(4/3)*pi*d(1)^3;

W(1)=V(1)*p;

while d(m)>.2;

ti=ti+1;

time(m+1)=ti;

A=pi*d(m)^2;

W(m+1)=W(m)-.2*A^.5;

V(m+1)=W(m+1)/p;

d(m+1)=(V(m+1)*(3/4)*(1/pi))^(1/3);

m=m+1;

end

plot(time,d,'k');xlabel('Time (day)')

ylabel('Sphere diameter (cm)')


43/44

Q21: Plot Pxy diagram for the Binary system of acetonitrile(1)/nitromethane(2). Vaporpressures for the pure species are given by the following equations.For acetonitrile(1) P

o1=exp(14.2724-2945.47/(T+224))

For nitromethane(2) Po2= exp(14.2043-2972.64/(T+209))In whichT=75 CoP1 and P2 in kpaKi= P

oi /Pt

At Bubble point yi=Kixi =1Solution:

Write the following codeX1=0:.1:1;

X2=1-X1;

T=75;

P1=exp(14.2724-2945.47/(T+224));% Vapor pressure of acetonitrile

P2=exp(14.2043-2972.64/(T+209));% Vapor pressure of nitromethane

for m=1:11

for Pt=.1:.01:120;

K1=P1/Pt; % acetonitrile


44/44

K2=P2/Pt; % nitromethane

sum=K1*X1(m)+K2*X2(m);

if sum

Documents

232541184 Certain Numerical Problems Chemical Engineering MATLAB